Difference between revisions of "1998 AIME Problems/Problem 14"

Revision as of 13:17, 3 May 2020

Problem

An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$ ?

Solution 1

$2mnp = (m+2)(n+2)(p+2)$

Let’s solve for $p$ :

$(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)$ $[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)$ $p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}$

Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$ . The possible pairs of factors of $9$ are $(1,9)(3,3)$ . These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We now check that $130$ is optimal, setting $a=m-2$ , $b=n-2$ in order to simplify calculations. Since $0 \le (a-1)(b-1) \implies a+b \le ab+1$ We have $p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130$ Where we see $(m,n)=(3,11)$ gives us our maximum value of $\boxed{130}$ .

Note that $0 \le (a-1)(b-1)$ assumes $m,n \ge 3$ , but this is clear as $\frac{2m}{m+2} = \frac{(n+2)(p+2)}{np} > 1$ and similarly for $n$ .

Solution 2

Similarly as above, we solve for $p,$ but we express the denominator differently:

$p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.$ Hence, it suffices to maximize $\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.

Then since $\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ $\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+2)}{5(n+2)}=\dfrac{n-10}{10(n+2)},$ where we simply let $n=11$ to achieve $p=\boxed{130}.$

~Generic_Username

Solution 3

Observe that $2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{2}{m} \right )^3$ thus $m < 7$ .

Now, we can use casework on $m$ and Simon's Favorite Factoring Trick to check that $m = 5,2,1$ have no solution and for $m = 3,4,6$ , we have the corresponding values of $p$ : $130,54,16$ .

Thus, the maximum value is $\boxed{130}$ .

~amplreneo

@@ Line 34: / Line 34: @@
 == Solution 3 ==
 Observe that
-<cmath>2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{1}{m} \right )^3</cmath> thus <math>m < 7</math>.
+<cmath>2 = \left ( 1 + \frac{2}{m} \right ) \left ( 1 + \frac{2}{n} \right ) \left (1 + \frac{2}{p} \right ) \leq \left ( 1 + \frac{2}{m} \right )^3</cmath> thus <math>m < 7</math>.
 Now, we can use casework on <math>m</math> and Simon's Favorite Factoring Trick to check that <math>m = 5,2,1</math> have no solution and for <math>m = 3,4,6</math>, we have the corresponding values of <math>p</math>: <math>130,54,16</math>.
@@ Line 41: / Line 41: @@
 ~amplreneo
 == See also ==
 {{AIME box|year=1998|num-b=13|num-a=15}}

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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